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The [H⁺] in a 1.0 M solution of Naâ‚‚CO₃ is (for Hâ‚‚CO₃, Ka₁ = 4.3 x 10⁻⁷; Kaâ‚‚ = 5.6 x 10⁻¹¹):

a) 7.5 x 10⁻⁶
b) 6.6 x 10⁻⁴
c) 1.3 x 10⁻²
d) 7.5 x 10⁻¹³
e) none of these

Answer :

Final answer:

To find the [H+] in a 1.0 M Na2CO3 solution, we use the hydrolysis constant Ki of bicarbonate, which leads us to the conclusion that [H+] is approximately equal to Ki. Hence, the hydrogen ion concentration in a 1.0 M solution of Na2CO3 is 6.6 x 10^-4 (option b).

Explanation:

To calculate the hydrogen ion concentration ([H+]) in a 1.0 M solution of Na2CO3, we need to consider the two dissociation steps of carbonic acid (H2CO3). The first dissociation has the constant Ka1 = 4.3 x 10−7 and the second dissociation has the constant Ka2 = 5.6 x 10−11. In aqueous solution, Na2CO3 completely dissociates into Na+ ions, which are inert, and CO32− ions, which will react with water to form HCO3− ions and OH− ions.

The main reaction affecting the pH is the hydrolysis of the bicarbonate ion (HCO3−), an amphiprotic species. It can donate a proton to produce CO32− or accept a proton to produce H2CO3. Because CO32− will be the predominant species in a 1 M Na2CO3 solution, we should consider that it can react with water, giving HCO3− and OH−. But, since we are interested in [H+] and not [OH−], we should look at the reverse reaction, where HCO3− accepts H+ from water, increasing the [H+].

Using the approximation that the dissociation of HCO3− into CO32− is negligible, the [H+] is governed by the expression for the hydrolysis constant Ki of bicarbonate which is Kw/Ka1, where Kw is the ion-product constant for water (1.0 x 10−14 at 25 °C).

This yields the equation Ki = [HCO3−][H+]/[CO32−], which allows us to solve for [H+]. Given that Ki = 2.3 x 10−8, we can set up the following:

[H+] = Ki x [CO32−] / [HCO3−]

Since the stoichiometry between CO32− and HCO3− is 1:1, [H+] approximately equals Ki. Therefore, the answer is (b) 6.6 x 10−4.

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Rewritten by : Jeany

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